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mathcalculus

Help: CONTINUITY and removable discontinuity: (attached below)

  • 8 months ago
  • 8 months ago

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  1. mathcalculus
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    • 8 months ago
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  2. mathcalculus
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    @kropot72

    • 8 months ago
  3. Psymon
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    So you have 2 conditions to have a removable discontinuity. The first one is simply that the value must not be defined at that point, which is pretty obvious that is isn't because of the denominator. The second condition is basically the difference between whether or not you have an asymptote at your undefined point or simply a hole. The way to tell whether or not its an asymptote or a whole is if you factored your function, would the factor that is undefined cancel or remain? If the factor on bottom cancels, it is merely a hole, if it does not cancel then it is an asymptote and CANNOT be removed. So I would factor your numerator and see if the denominator cancels. If it does, then simply plug in 4 into your result and you'll get the value you're looking for.

    • 8 months ago
  4. mathcalculus
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    can we work on this together?

    • 8 months ago
  5. mathcalculus
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    i've tried doing this on my own but seem to be getting something wrong.

    • 8 months ago
  6. Psymon
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    Sure. So I would factor the numerator and see if x-4 cancels. If it cancels, you have a hole, if it does not cancel, you have an asymptote and it cannot be removed. You know how to factor the numerator, right?

    • 8 months ago
  7. mathcalculus
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    let me try.

    • 8 months ago
  8. Psymon
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    Alright, sure.

    • 8 months ago
  9. mathcalculus
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    i got 2(x^2+3x-25) @Psymon

    • 8 months ago
  10. Psymon
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    -28you mean? And after that, you still have to further factor the x^2 + 3x - 28.

    • 8 months ago
  11. Psymon
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    Find factors of -28 that will add up to positive 3.

    • 8 months ago
  12. mathcalculus
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    yes, ,-28

    • 8 months ago
  13. Psymon
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    Right. So When you further factor the numerator, you should have 2(x±__)(x±__). Its just remember how to factor that x^2 + 3x - 28 portion.

    • 8 months ago
  14. mathcalculus
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    2(x+7) (x-4)

    • 8 months ago
  15. Psymon
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    Exactly. In turn, that cancels out the denominator, which means the discontinuity is removable and not just an asymptote. Not only that, since the term that would make your function undefined disappeared, you can simply plug in 4 to find out the value you need.

    • 8 months ago
  16. mathcalculus
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    we're left with 2 (x+7)

    • 8 months ago
  17. mathcalculus
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    so we plug in 4?

    • 8 months ago
  18. Psymon
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    That's all ya do.

    • 8 months ago
  19. mathcalculus
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    does this apply to every problem with continuity?

    • 8 months ago
  20. Psymon
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    Yes. Most of the work is finding a way to eliminate the portion that would give you an undefined answer. Once you manage that, you just plug in the value that the limit is approaching. Of course there are times when you cannot remove the undefined portion, in which that means the limit does not exist and you are merely approaching an asymptote.

    • 8 months ago
  21. mathcalculus
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    thanks:)

    • 8 months ago
  22. Psymon
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    Sure :3

    • 8 months ago
  23. mathcalculus
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    ohhh can i practice a little, then ask you for a tricky problem like the one you just mentioned? it would really help.

    • 8 months ago
  24. Psymon
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    That works, I have no problem with that.

    • 8 months ago
  25. mathcalculus
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    what about a JUMP DISCONTINUITY? same thing?

    • 8 months ago
  26. Psymon
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    That's new terminology for me. I probably know what it is but have not heard it said that way. Could you give me an example?

    • 8 months ago
  27. mathcalculus
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    sure

    • 8 months ago
  28. mathcalculus
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    1 moment

    • 8 months ago
  29. mathcalculus
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    • 8 months ago
  30. Psymon
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    Ah, just one sided limits. We never really gave a name for it before, just that it had to do with one-sided, lol.

    • 8 months ago
  31. mathcalculus
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    oh so how do we begin with this one? sketch a graph?

    • 8 months ago
  32. mathcalculus
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    i can do graphs but i'd like to do it a simpler way since i won't have enough time when it comes to a final

    • 8 months ago
  33. mathcalculus
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    @Psymon

    • 8 months ago
  34. Psymon
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    I apologize, I guess I was making sure what jump kind of meant since I did not have a picture. But yes, all you really need to confirm is that when x = 6 on both functions that they have different limits. It is pretty obvious that they do. As far as the left and right part goes, you need to see which function to just check and see which function is approaching from the left and which one is approaching from the right. Since 8x-8 < 6, that must be the function approaching from the left. Therefore, f(6) of 8x-8 is the value you will use for limit from the left. Since the other function is the one that will becoming from the right, it's value at x = 6 will be the value you use for the right-sided limit. Basically, you just need to see which function to plug the value into.

    • 8 months ago
  35. Psymon
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    The delay was graphing and looking at so I knew exactly what to say and to get my brain to work, lol.

    • 8 months ago
  36. mathcalculus
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    it's okay, so how can we work this prob out?

    • 8 months ago
  37. Psymon
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    |dw:1375046904734:dw| I know you know how to graph, but I wanted the visual up before we move on. So you just want to see which function is from the left and which one is from the right. When it gives you a piece-wise function, it's pretty obvious which one is going to be coming from the left and which one from the right. Once you determine that, just plug in x = 6 into the graph that comes from the left and use that value for lim ->6 - Once you do that, plug x = 6 into the other function, since you know it's the one on the right, and find lim x-> 6 +

    • 8 months ago
  38. Psymon
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    A lot of the time you may have to determine the behavior on your own, but the piece-wise just gave you the behavior so you didn't have to worry about it. From there it was just picking the correct function for left and for right.

    • 8 months ago
  39. Psymon
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    In regards to removable or non-removable,it works the same way. if one of the functions has an asymptote at the jump point, then one of the sides or both may have a limit that DNE on that side, left or right. Maybe another example?

    • 8 months ago
  40. mathcalculus
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    so to this problem we HAVE to sketch only?

    • 8 months ago
  41. Psymon
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    In this case, no, because the piece-wise function tells us which function to use on the left and on the right. The sketching would only come in if you had maybe one function that wasn't a piece-wise. Like if the function was 1/[(2x^2 -3] For something like that you would have to test points on the left and right and see if they're going up to infinity, down to negative infinity, etc. Any of these piece-wise functions won't require s ketch, I just wanted to draw visual aid.

    • 8 months ago
  42. Psymon
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    Holy cow my typing and grammar are horrible today o.o

    • 8 months ago
  43. Psymon
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    Maybe if you have another jump problem we can look at it.

    • 8 months ago
  44. Psymon
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    I'm going to need to head out, I apologize. Good luck with the limits.

    • 8 months ago
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